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June 21st, 2014, 05:15 PM  #1 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Which side is greater? my challenge #2
$\displaystyle \sqrt{2} \ + \ \sqrt{3} \ + \ \sqrt{5} \ + \ \sqrt{7} \ \ \ versus \ \ \ 8 $ I am not asking to do the actual work to determine which side is larger. The question is: At any given step, is it possible to eliminate one of the four radicals that will be present in that step after squaring, adding, and/or subtracting? If that can be done, then the original problem is possible by my method illustrated here: Which side is greater? my challenge #1 post #3 Last edited by Math Message Board tutor; June 21st, 2014 at 05:19 PM. 
June 21st, 2014, 06:08 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 Math Focus: Mainly analysis and algebra 
If you get two roots only on one side and then square, they will become one after squaring.

June 22nd, 2014, 07:45 AM  #3 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  
June 22nd, 2014, 08:51 AM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
$$ \sqrt2+\sqrt3>^?8\sqrt5\sqrt7 $$ and square you do have only one radical on the left, but you have three on the right: $$ 5+2\sqrt6>^?7616\sqrt516\sqrt7+2\sqrt{35} $$ Now fortunately that's not a problem here since you can transpose and square again to get $$ 24>^?82532720\sqrt52592\sqrt7+796\sqrt{35} $$ then move the negatives around (and fold the constant) to get $$ 2720\sqrt5+2592\sqrt7>^?8229+796\sqrt{35} $$ which can be squared to get one radical on each side and solved by (transposing and squaring) twice. But this isn't general.  

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